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Let $\lambda_1, \lambda_2, \lambda_3$ denote the eigenvalues of the matrix $ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos  t & \sin  t \\ 0 & -\sin  t & \cos  t \end{pmatrix}$.

If $\lambda_1+ \lambda_2+\lambda_3=\sqrt{2} +1$ then the set of possible values of $t, - \pi \leq t < \pi$, is

  1. Empty set
  2. $\{ \frac{\pi}{4} \}$
  3. $\{ - \frac{\pi}{4}, \frac{\pi}{4} \}$
  4. $\{ - \frac{\pi}{3}, \frac{\pi}{3} \}$

 

asked in Linear Algebra by Loyal (3.6k points)  
edited by | 66 views

1 Answer

+1 vote

Answer C)

 

1-⋋  0        0

0    cos t-⋋ sin t

0     -sin t    cos t - ⋋

 

 

solving (1-⋋) {(cos t - ⋋)2 - sin2⋋}=0................i

⋋=1

⋋1+⋋2+⋋3=1+√2

if ⋋1=1

⋋2+⋋3=√2

 

from equation i we can say (cos t - ⋋)2 - sin2⋋=0

(cos t - ⋋) =(+/-) sin⋋

⋋= cos t -sin t

t=(⊼/2⨉2 + ⊼/4)  as range of t is -⊼<t<⊼

So, ans should be C)

answered by Veteran (52.4k points)  


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